Question

$\int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx$ is equal to

• A. $\frac{1}{2}\sin 2x+C$
• B. $-\frac{1}{2}\sin 2x+C$
• C. $-\frac{1}{2}\sin x+C$
• D. $-{\sin }^{2}x+C$

Answer 1

The correct option is B $-\frac{1}{2}\sin 2x+C$

$\int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx$

$\Rightarrow$ $\int \frac{({\sin }^{4}x+{\cos }^{4}x)({\sin }^{4}x-{\cos }^{4}x)}{({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x}dx$

$\Rightarrow$ $\int \frac{({\sin }^{4}x+{\cos }^{4}x)({\sin }^{4}x-{\cos }^{4}x)}{{\sin }^{4}x+{\cos }^{4}x}dx$

$\Rightarrow$ $\int ({\sin }^{4}x-{\cos }^{4}x)dx$

$\Rightarrow$ $\int ({\sin }^{2}x-{\cos }^{2}x)({\sin }^{2}x+{\cos }^{2}x)dx$

$\Rightarrow$ $\int ({\sin }^{2}x-{\cos }^{2}x)dx$

$\Rightarrow$ $-\int \cos 2xdx$

$\Rightarrow$ $-\frac{1}{2}\sin 2x+c$